In the triangle ABC on the side BC, the pointsD and E are chosen so that the angle BAD is equal to the angle EAC. Let I and J be the centers of the inscribed circles of triangles ABD and AEC respectively, F be the point of intersection of BI and EJ, G be the point of intersection of DI and CJ. Prove that the points I,J,F,G lie on one circle, the center of which belongs to the line IbIc, where Ib and Ic are the centers of the exscribed circles of the triangle ABC, which touch respectively sides AC and AB. geometryConcyclicexcenterUkrainian TYM